Cartesian Product is Small
 It has been suggested that this page or section be merged into Equivalence of Definitions of Ordered Pair. (Discuss) |
Theorem
Let $a$ and $b$ be small classes.
Then their Cartesian product $\left({a \times b}\right)$ is small:
- $\mathscr M \left({a \times b}\right)$
Proof
| \(\displaystyle \left({a \times b}\right)\) | \(=\) | \(\displaystyle \left\{ {\left({x, y}\right) : x \in a \land y \in b}\right\}\) | $\quad$ Definition of Cartesian Product | $\quad$ | |||||||||
| \(\displaystyle \) | \(=\) | \(\displaystyle \left\{ {\left({x, y}\right) : \left\{ {x}\right\} \subseteq a \land \left\{ {y}\right\} \subseteq b} \right\}\) | $\quad$ Singleton of Element is Subset | $\quad$ | |||||||||
| \(\displaystyle \) | \(\subseteq\) | \(\displaystyle \left\{ {\left({x, y}\right) : \left\{ {x, y}\right\} \subseteq \left({a \cup b}\right)}\right\}\) | $\quad$ Set Union Preserves Subsets | $\quad$ | |||||||||
| \(\displaystyle \) | \(\subseteq\) | \(\displaystyle \left\{ {\left({x, y}\right) : \left({x, y}\right) \subseteq \mathcal P \left({a \cup b}\right)}\right\}\) | $\quad$ Power Set of Subset and Subset Relation is Transitive | $\quad$ |
It's not obvious how $\left({x, y}\right) \subseteq \mathcal P \left({a \cup b}\right)$
You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by explaining it. To discuss this point in more detail, feel free to use the talk page.
If you are able to explain it, then when you have done so you can remove this instance of {{Explain}} from the code.
So by the definition of power set:
- $\left({a \times b}\right) \subseteq \mathcal P \left({\mathcal P \left({a \cup b}\right)}\right)$
By Union of Small Classes is Small, $\left({a \cup b}\right)$ is small.
By the Axiom of Powers, $\mathcal P \left({\mathcal P \left({a \cup b}\right)}\right)$ is small.
By Axiom of Subsets Equivalents, $\left({a \times b}\right)$ is small.
$\blacksquare$
Sources
- 1971: Gaisi Takeuti and Wilson M. Zaring: Introduction to Axiomatic Set Theory: $\S 6.2$
